Pearson Algebra 2 Common Core, 2011
PA
Pearson Algebra 2 Common Core, 2011 View details
5. Systems With Three Variables
Continue to next subchapter

Exercise 24 Page 172

When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable.

(5,2,2)

Practice makes perfect
The given system consists of equations of planes. When using the Substitution Method to solve a system of equations, it is necessary to isolate a variable. In the third equation, z is already isolated, so we will start by substituting its value into the second equation.
13=3x-y 4y-3x+2z=-3 z=2x-4y
13=3x-y 4y-3x+2( 2x-4y)=-3 z=2x-4y
â–¼
(II): Solve for x
13=3x-y 4y-3x+4x-8y=-3 z=2x-4y
13=3x-y x-4y=-3 z=2x-4y
13=3x-y x=4y-3 z=2x-4y
This time, the x-variable was isolated in the second equation. We can now substitute its equivalent expression into the first equation.
13=3x-y x=4y-3 z=2x-4y
13=3( 4y-3)-y x=4y-3 z=2x-4y
â–¼
(I): Solve for y
13=12y-9-y x=4y-3 z=2x-4y
13=11y-9 x=4y-3 z=2x-4y
22=11y x=4y-3 z=2x-4y
2=y x=4y-3 z=2x-4y
y=2 x=4y-3 z=2x-4y
The value of y is 2. Substituting 2 for y into the second equation, we can find the value of x.
y=2 x=4y-3 z=2x-4y
y=2 x=4( 2)-3 z=2x-4y
y=2 x=8-3 z=2x-4y
y=2 x=5 z=2x-4y
Now that we know the values of x and y, we are able to find the value of z.
y=2 x=5 z=2x-4y
y=2 x=5 z=2( 5)-4( 2)
â–¼
(III): Evaluate right-hand side
y=2 x=5 z=10-8
y=2 x=5 z=2
The solution to the system is the point ( 5, 2, 2). This is the singular point at which all three planes intersect. Let's check our solution by substituting the values into the system.
13=3x-y 4y-3x+2z=-3 z=2x-4y

(I), (II), (III): Substitute values

13? =3( 5)- 2 4( 2)-3( 5)+2( 2)? =-3 2? =2( 5)-4( 2)

(I), (II), (III): Multiply

13? =15-2 8-15+4? =-3 2? =10-8

(I), (II), (III): Add and subtract terms

13=13 ✓ -3=-3 ✓ 2=2 ✓
Since the substitution of our answers into the given equations resulted in three identities, we know that our solution is correct.